$AD$, $BE$ and $CF$ are concurrent lines on $\triangle ABC$. Show that lines through the midpoints of $BC$, $CA$ and $AB$ respectively parallel to $AD$, $BE$ and $CF$ are concurrent.
Let $M_1$, $M_2$ and $M_3$ be midpoints of $BC$, $AC$ and $AB$ respectively. Lines parallel to $AD$ and $BE$ pass through $M_1$ and $M_2$ respectively meet at point $O$. It suffices to prove that line from $M_3$ to $O$ is parallel to $CF$.
I am guessing that this has to do with Ceva's theorem. But I am unable to use it to prove the statement. It is really confusing because of so many lines in the figure. Kindly post your solutions.
Best Answer
The original triangle $ABC$ and its medial triangle $M_1 M_2 M_3$ are related via a dilation $\delta$ centered at the common centroid $G$. Any dilation preserves parallel lines and concurrency, so if $AD,BE,CF$ are concurrent so they are their images via $\delta$, i.e. the given lines through $M_1,M_2,M_3$.